I'm with a problem in an exercise form Do Carmo's Differential Geometry of Curves and Surfaces; it is number 9 section 1.6.

I have a differentiable real function $k(s)$, $s\in I$, and I need to show that the planar curve given by
$$
\alpha (s)=\left( \int \cos\theta (s)ds +a,\int \sin\theta (s)ds +b\right)
$$
where
$$
\theta (s) = \int k(s)ds+\phi
$$
has $k(s)$ as its curvature (there is an explanation for this curvature not being exclusively positive in the same book). Both a,b and $\phi$ are constants so they are only important for the second part of the exercise (show that the curve is determined uniquely except of translations and rotations by a $\phi$-angle).

So what I need to show that the curvature of $\alpha$ is $k(s)$. I have already solved exercise 12.d from the same section of the book which gives me

This also gives me $|\alpha '(s)|= |k(s)|$. Substituting on the formula for $k(s)$ above (which should give me k(s)=k(s) ) returns me the value
$$
\dfrac{k(s)}{|k(s)|}
$$
which is $\pm$ and also very frustrating (shouldn't end up with any restrictions over $k$ ). I can't find where I'm making the mistake. I have tried to do this with arc lenght parametrization, which gave me the same result (by the way, that's how I obtained the formula for $k(s)$).

The derivate of $\int\cos(\theta(s))$ is simply $\cos(\theta(s))$. So you should find $|\alpha'(s)|=1$.
–
XoffAug 28 '12 at 18:23

... and therefore we have an example of a really simple problem that got complicated because of a wrong usage of the Calculus Fundamental Theorem. Thanks @Xoff ! That was is accurate and correct.
–
MarraAug 28 '12 at 18:33